Igor is essentially right. If $n > 2$ and the embedding is convex, then the second fundamental form exists almost everywhere. Using the Gauss equations per Robert's answer, it extends uniquely and smoothly to everywhere. The rest is straightforward. I'm omitting it because I'm typing this on an iPhone.
The $C^1$ but non-convex case is due to Nash (who did it in codimension 2) and Kuiper (who did it in codimension 1). It is not a special case of smooth Nash embedding theorem.
ADDED: The argument can be made rigorous by taking a family of smooth convex hypersurfaces converging uniformly to the original convex hypersurface and using the inverse to the map from positive definite second fundamental forms to curvature tenors defined by the Gauss equations to show that the second fundamental form has to converge uniformly to a continuos limit that also weakly solves the Codazzi equations. This can then be integrated to show that the embedding is in fact smooth. Uniqueness then follows by the Cohn-Vossen theorem.